Union & Intersection for Propositional Logic How is it that the union of two sets of sentences (that, individually, logically entail a sentence) logically entails a sentence while the intersection of the two sets does not logically entail a sentence?

 A: *

*If it is raining, then my car is wet.

*It is raining.

*If I am in the carwash, then my car is wet. 

*I am in the carwash.
Set $S$ to be the first two propositions, $T$ to be the last two.  Each of $S,T$ imply that my car is wet.  Their union, i.e. all four propositions above, also implies that my car is wet.  However $S\cap T=\emptyset$, which does not imply that my car is wet.
A: First of all, I have only been studying logic for a week, so feel free to correct me.
For your first question, let the two sets be Γ = {a,b} and Δ = {c,d} with a,b,c and d being propositional sentences (forgive me for being lazy); then Γ ⊨ φ tells us a∧b ⇒ φ, and Δ ⊨ φ means c∧d ⇒ φ. As vadim says above, Γ ∩ Δ = ∅ (empty set). Recall the definition of entailment: for Γ ∩ Δ to entail φ, every truth assignment that satisfies Γ ∩ Δ should also satisfy φ. Now, any truth assignment satisfies an empty set, so φ should hold under any truth assignment, i.e. φ is a tautology. But in our context no such information is mentioned, so the first proposition is false.
For the second question Γ ⊨ φ and Δ ⊭ φ, then Γ ∪ Δ ⊨ φ. WLOG, this time let Γ = {a,b} and Δ = {b,c}. Γ ⊨ φ means any truth assignment that satisfies a and b also satisfy φ, i.e. a∧b ⇒ φ. Δ ⊭ φ means some truth assignment that satisfies b and c does not satisfy φ. Now Γ ∪ Δ = {a,b,c}, so the question is about whether a∧b∧c ⇒ φ. Notice that any truth assignment that satisfies a∧b∧c must satisfy a∧b as well, so it also satisfies φ, and we have that Γ ∪ Δ entails φ. The fact that Δ ⊭ φ has nothing to do with it, since we know φ must be true as long as a∧b is true.
IMO, you just need to be aware that Γ ∪ Δ ≠ Γ ∨ Δ - this seems to be an easy mistake.
